$n_{Fe}=\dfrac{5,6}{56}=0,1 mol$
$Fe+2H^+\to Fe^{2+}+H_2$
$\Rightarrow n_{H^+}=0,1.2=0,2 mol$
$n_{H^+}=n_{HCl}+2n_{H_2SO_4}=0,5V_1+2V_1=2,5V_1(mol)$
$\Rightarrow 2,5V_1=0,2\Leftrightarrow V_1=0,08(l)$
$n_{H_2}=n_{Fe}=0,1 mol$
$\Rightarrow V_2=0,1.22,4=2,24l$